Question: Divide the following complex numbers. $ \dfrac{-6+8i}{-4-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+3i}$ $ \dfrac{-6+8i}{-4-3i} = \dfrac{-6+8i}{-4-3i} \cdot \dfrac{{-4+3i}}{{-4+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-6+8i) \cdot (-4+3i)} {(-4-3i) \cdot (-4+3i)} = \dfrac{(-6+8i) \cdot (-4+3i)} {(-4)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-6+8i) \cdot (-4+3i)} {(-4)^2 - (-3i)^2} = $ $ \dfrac{(-6+8i) \cdot (-4+3i)} {16 + 9} = $ $ \dfrac{(-6+8i) \cdot (-4+3i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-6+8i}) \cdot ({-4+3i})} {25} = $ $ \dfrac{{-6} \cdot {(-4)} + {8} \cdot {(-4) i} + {-6} \cdot {3 i} + {8} \cdot {3 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{24 - 32i - 18i + 24 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{24 - 32i - 18i - 24} {25} = \dfrac{0 - 50i} {25} = -2i $